The Stacks project

Proof. The equivalence holds because the support of an element of $M$ is contained in the support of $M$ and conversely the support of $M$ is the union of the supports of its elements. The category of these modules is a Serre subcategory (Homology, Definition 12.10.1) of $\text{Mod}_ A$ by Algebra, Lemma 10.40.9. We omit the proof of the statement on direct sums. $\square$

Proof. This follows from the fact that the functor $H^0_ T(-)$ is the right adjoint to the inclusion functor $\text{Mod}_{A, T} \to \text{Mod}_ A$, see Derived Categories, Lemma 13.30.3. $\square$

Proof. Let $J^\bullet $ be a K-injective complex representing $K$. By definition $RH^0_ T$ is represented by the complex

where the equality follows from our definition of $H^0_ T$. Since filtered colimits are exact the cohomology of this complex in degree $i$ is $\mathop{\mathrm{colim}}\nolimits H^ i(H^0_ Z(J^\bullet )) = \mathop{\mathrm{colim}}\nolimits H^ i_ Z(K)$ as desired. $\square$

Proof. Let $M$ be an object of $\text{Mod}_{A, T}$. Choose an embedding $M \to J$ into an injective $A$-module. By Dualizing Complexes, Proposition 47.5.9 the module $J$ is a direct sum of injective hulls of residue fields. Let $E$ be an injective hull of the residue field of $\mathfrak p$. Since $E$ is $\mathfrak p$-power torsion we see that $H^0_ T(E) = 0$ if $\mathfrak p \not\in T$ and $H^0_ T(E) = E$ if $\mathfrak p \in T$. Thus $H^0_ T(J)$ is injective as a direct sum of injective hulls (by the proposition) and we have an embedding $M \to H^0_ T(J)$. Thus every object $M$ of $\text{Mod}_{A, T}$ has an injective resolution $M \to J^\bullet $ with $J^ n$ also in $\text{Mod}_{A, T}$. It follows that $RH^0_ T(M) = M$.

Next, suppose that $K \in D_ T^+(A)$. Then the spectral sequence

(Derived Categories, Lemma 13.21.3) converges and above we have seen that only the terms with $q = 0$ are nonzero. Thus we see that $RH^0_ T(K) \to K$ is an isomorphism. Thus the functor $D^+(\text{Mod}_{A, T}) \to D^+_ T(A)$ is an equivalence with quasi-inverse given by $RH^0_ T$. $\square$

Proof. Say $\dim (A) = d$. Then we see that $H^ i_ Z(M) = 0$ for $i > d$ for every closed subset $Z$ of $\mathop{\mathrm{Spec}}(A)$, see Lemma 51.4.7. By Lemma 51.5.3 we find that $H^0_ T$ has bounded cohomological dimension.

Let $K \in D_ T(A)$. We claim that $RH^0_ T(K) \to K$ is an isomorphism. We know this is true when $K$ is bounded below, see Lemma 51.5.4. However, since $H^0_ T$ has bounded cohomological dimension, we see that the $i$th cohomology of $RH_ T^0(K)$ only depends on $\tau _{\geq -d + i}K$ and we conclude. Thus $D(\text{Mod}_{A, T}) \to D_ T(A)$ is an equivalence with quasi-inverse $RH^0_ T$. $\square$

Proof. From the map $R\Gamma _ T(K) \to K$ we get a map $R\Gamma _ T(K) \otimes _ A^\mathbf {L} B \to K \otimes _ A^\mathbf {L} B$. The cohomology modules of $R\Gamma _ T(K) \otimes _ A^\mathbf {L} B$ are supported on $T'$ and hence we get the arrow of the lemma. This arrow is an isomorphism if $T$ is a closed subset of $\mathop{\mathrm{Spec}}(A)$ by Dualizing Complexes, Lemma 47.9.3. Recall that $H^ i_ T(K)$ is the colimit of $H^ i_ Z(K)$ where $Z$ runs over the (directed set of) closed subsets of $T$, see Lemma 51.5.3. Correspondingly $H^ i_{T'}(K \otimes _ A^\mathbf {L} B) = \mathop{\mathrm{colim}}\nolimits H^ i_{Z'}(K \otimes _ A^\mathbf {L} B)$ where $Z'$ is the inverse image of $Z$. Thus the result because $\otimes _ A B$ commutes with filtered colimits and there are no higher Tors. $\square$

Proof. Let $E$ be an object of $D_{T \cap T'}(A)$. Then we have

The first equality by the adjointness property of $R\Gamma _ T$ and the second by the adjointness property of $R\Gamma _{T'}$. On the other hand, if $J^\bullet $ is a bounded below complex of injectives representing $K$, then $H^0_{T'}(J^\bullet )$ is a complex of injective $A$-modules representing $R\Gamma _{T'}(K)$ and hence $H^0_ T(H^0_{T'}(J^\bullet ))$ is a complex representing $R\Gamma _ T(R\Gamma _{T'}(K))$. Thus $R\Gamma _ T(R\Gamma _{T'}(K))$ is an object of $D^+_{T \cap T'}(A)$. Combining these two facts we find that $R\Gamma _{T \cap T'} = R\Gamma _ T \circ R\Gamma _{T'}$. This produces the spectral sequence by the lemma referenced in the statement. $\square$

Proof. By construction we may represent $R\Gamma _ T(A)$ by a complex $J^\bullet $ in $\text{Mod}_{A, T}$. Thus if we represent $K$ by a K-flat complex $K^\bullet $ then we see that $R\Gamma _ T(A) \otimes _ A^\mathbf {L} K$ is represented by the complex $\text{Tot}(J^\bullet \otimes _ A K^\bullet )$ in $\text{Mod}_{A, T}$. Using the map $R\Gamma _ T(A) \to A$ we obtain a map $R\Gamma _ T(A) \otimes _ A^\mathbf {L} K\to K$. Thus by the adjointness property of $R\Gamma _ T$ we obtain a canonical map

factoring the just constructed map. Observe that $R\Gamma _ T$ commutes with direct sums in $D(A)$ for example by Lemma 51.5.3, the fact that directed colimits commute with direct sums, and the fact that usual local cohomology commutes with direct sums (for example by Dualizing Complexes, Lemma 47.9.1). Thus by More on Algebra, Remark 15.59.11 it suffices to check the map is an isomorphism for $K = A[k]$ where $k \in \mathbf{Z}$. This is clear.

The final statements follow from the result we've just shown by transitivity of derived tensor products. $\square$